Consider the additive group of integer sequences $\mathbb{Z}^{\mathbb{N}}$. Why does every epimorphism of groups $\mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ split? $(\star)$

Actually this claim is equivalent to the Whitehead problem for countable abelian groups:

"$\Rightarrow$": Recall Specker's result which states that $\mathbb{Z}^{(\mathbb{N})} \to \hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$, $e_i \mapsto \mathrm{pr}_i$ is an isomorphism, i.e. $\mathbb{Z}^{\oplus \mathbb{N}}$ is reflexive. Now assume that $A$ is countable and $\mathrm{Ext}^1(A,\mathbb{Z})=0$. Choose a presentation $0 \to P \to Q \to A \to 0$ with free abelian groups $P,Q$, w.l.o.g. of rank $\aleph_0$. By assumption $Q^* \to P^*$ is an epi, hence splits. Since $P,Q$ are reflexive, then also $P \to Q$ splits, and $A$ is free.

"$\Leftarrow$": If $f : \mathbb{Z}^{\mathbb{N}} \to \mathbb{Z}^{\mathbb{N}}$ is an epimorphism, then $f^* : \mathbb{Z}^{\oplus \mathbb{N}} \to \mathbb{Z}^{\oplus \mathbb{N}}$ is a monomorphism, the cokernel $A$ of $f^*$ is countable and satisfies $\mathrm{Ext}^1(A,\mathbb{Z})=0$, since $f^{**} \cong f$ is epi. Hence $A$ is free, which implies that $f^*$ splits and therefore also $f^{**} \cong f$ splits. $~\square$

The countable Whitehead problem was proved by Stein in 1950. He used injective resolutions, i.e. $\mathrm{Ext}^1(A,\mathbb{Z}) = \mathrm{coker}(\hom(A,\mathbb{Q}) \to \hom(A,\mathbb{Q}/\mathbb{Z}))$. In particular, $(\star)$ is true. On the other hand, the equivalence above suggests an alternative proof of the countable Whitehead problem. Therefore my question is: Is there a direct proof for $(\star)$?

By Specker's result an endomorphism of $\mathbb{Z}^{\mathbb{N}}$ corresponds to an endomorphism of $\mathbb{Z}^{\oplus \mathbb{N}}$, and therefore to a column-finite matrix. But I don't know how to characterize surjectivity. And this is where I get stuck.